Final answer:
The question involves setting up and solving a differential equation to find the amount of salt in a tank over time, taking into account the rates of addition and drainage of a saline solution.
Step-by-step explanation:
The amount of salt in the tank as a function of time can be derived using the concept of differential equations. Initially, the tank contains 100 L of water without any salt. As the saline solution with a concentration of 0.5 kg/L is added at a rate of 6 L/min, salt begins to accumulate. Also, the tank is being drained at a rate of 4 L/min, which removes both water and dissolved salt.
We can write an equation that represents the rate of change of the amount of salt in the tank (S) at any time (t):
dS/dt = rate in - rate out
Rate in is the concentration of the incoming solution (0.5 kg/L) times the rate it is being added (6 L/min), so it's 0.5 * 6 = 3 kg/min. Rate out is the concentration of the salt in the tank at any time (S/100) times the rate at which the solution is being drained (4 L/min), so it's (S/100) * 4.
Combining these and assuming perfect mixing gives us the differential equation:
dS/dt = 3 - (4/100)S
By solving this equation, one can find the amount of salt in the tank as a function of time, which will typically involve integrating this expression and applying initial conditions.
Remembering that the initial amount of salt is zero, as there was only water in the tank at the beginning, will be essential in solving this problem to determine the function that describes the amount of salt (in kg) with respect to time (in minutes).